Live Sound University Article Sat, September 06, 2008

Hi -

I know there´s a lot of Smaart experts here so here goes a thought.

When using the delay locator, Smaart assumes that the right delay setting is at the highest peak of the waveform.

At low frequencies - is that really the right thing to do, I mean the “waveform” starts at a zero energy level..or? At 40 hz, there is some time between zero and top of the impulse. (Does Smaart use the same peak as the reference?)

I experimented with it, and did a measurement of a 2 way crossover going directly into Smaart with no delay applied. Smaart wanted to delay the low output. Why?

I also couldn´t get a coherent phase response with no delay on the low output. How does smaart analyse the phase response?

I also wonder what the difference is between 0 deg and 360 deg, is it only time dependent?

Thanks för your help
Patrik

Reply posted by Paul Tucci on September 17, 2001
- “When using the delay locator, Smaart assumes that the right delay setting is at the highest peak of the waveform.”

For an Impulse Alignment, yes, the peak is the right thing to use. For a smooth phase through acoustic crossover area alignment, it may NOT be the right timing. There are choices to be made about what to do with the info Smaart gives us.

- “At low frequencies, is that really the right thing to do? I mean the “waveform” starts at a zero energy level..or?” At 40 hz there is some time between zero and top of the impulse.”

Correct, and insightful.

- “I experimented with it, and did a measurement of a 2 way crossover going directly into Smaart with no delay applied. Smaart wanted to delay the low output. Why?”

I did this same experiment and scratched my head for months. Its called group delay. The low pass filter in the crossover causes some things to happen. Higher frequencies are rolled off at a chosen db/octave at a certain frequency. That’s the understandable part.

The PHASE shift caused by the crossover, and this is just a fact of life one cannot avoid, is problematic. Yes, it is in fact delaying the low frequency output. That’s why you measured it that way and that’s why Smaart’s phase trace was incoherent until you added the proper delay to synchronize your measurement.

Smaart gives you some good clues to tell you when measurements are bad. I am parent on duty this Am so I must run. I’m glad to see all these Smaart questions coming up now. I suggest to the powers that be that a Smaart-centered live chat is in order. I’ll elaborate more later.

PT

Reply posted by Patrik Arnekvist on September 17, 2001

Thanks for your response.
The thing is also, when I delayed the low output , I got a deep cancellation at the crossover point, if it all were in phase, why would that happen?

Patrik

Reply posted by Paul Tucci on September 18, 2001
-"The thing is also, when I delayed the low output , I got a deep cancellation at the crossover point, if it all were in phase, why would that happen?”

Your choice of wording is curious. You say Smaart wanted to delay the low end. Do you mean to say Smaart measured a time difference between the reference side of your equation and the output of your xover? That would be normal.

The delay would go on the reference side because you want to have compare the two when they are synchronous. “Row, row, row, your boat” vs “gently down the stream” does not a valid comparison make. Understand the analogy?

If you introduced delay onto the measurement side of the equation, ie, the low xover out, you made an even more invalid comparison. This would explain your lack of coherence. You may have created the transversal equalizer responsible for the notch described by comparing the same signal to itself, though one is offset in time at twice the wave period of your notch frequency. That’s a mouthful.

Then again, if you didn’t include the throughput delay of the digital (?) xover (approximately one millisecond)in your measurement, a 500 hz notch (1/2 of 1000hz or one millisecond of time) would appear. Could go either way.

PT

Reply posted by Earl Driggers on September 17, 2001
Keeping in mind that phase is time and time is phase, by delaying the low output the relative phase between the two outputs is probably 180^( or close ) now.

Another thing to keep in mind is that phase/time misallignment will only effect the frequencies common to both outputs. Now that the delay time is correct on the outputs, having to invert the phase on one of the outputs is a common occurance.

Reply posted by Chip on September 18, 2001
Earl,

I’d take some degree of exception to the phrase “phase is time, time is phase”. While this is essentially true, I submit that there can be differences.

For example, It’s entirely possible to have two completely time coherant signals. It’s just as possible to adjust the phase of one of those signals, without adjusting it’s time. Agreed?

I’m still trying to completely get my head around this one. But to the best of my understanding, the real trick seems to be realizing the difference. Here’s what I’ve found to be the marker between the two .... phase angle. If the time is correct, the phase angle of the two sources will be parallel. if the phase angle is leaning forward, or backward, it would indicate early or late arrival. In the case where the phase angle is parallel but they don’t overlap, phase adjustment would be the appropriate tool to slide the phase angle without changing the angle. Is this making sense?

Any input would be helpfull.

Chip

Reply posted by Phill Graham on September 18, 2001
Alright Chip, see if this helps. For a wave to be a wave, it has to propagate in both space and time, otherwise it is simply a vibration.

Math people wanted to be able describe things that change in a periodic manner. It turns out that trigonometric functions fit the bill nicely. However, you need a conversion factor that takes real world wave position and time, and translates it into the mathematical notion of “angles”.

Hidden in the equations for the wave propagation are two basic parameters. One basically says “swallow up” 30deg. of angle for every foot of travel in the air, and the other says “swallow up” 30 deg. of angle for ever .5sec of time elapsed (or whatever). The combination of these describes the wave’s movent in space and time. “Phase” lumps these two together.

It is important to realize that pure delay is not a flat (zero slope) phase plot, but rather a linear one with a slope proportional to the delay through whatever the device in question is. That is where the term “group delay” (-1st derivative [slope] of the phase) derives from, as the the group delay curve will be completely flat for an device with pure delay.

When Meyer sound talks about their phase correction through the spectrum, it is important to remember the small print at the bottom “phase vs. pure delay.” They are not correcting the phase to represent a flat (zero slope) line, but rather trying to get it to match the straight (but sloped) line of the pure delay between the speaker and the measurement system.

Let me know if this makes sense, I wrote it in a hurry.

Phill Graham

Reply posted by P.Tucci on September 18, 2001
-” I’d take some degree of exception to the phrase “phase is time, time is phase”.

Me too. While limiting your discussion to sine waves, I believe that to be true. A 90 degree phase shift of an 1120 hz tone is a quarter mSec of time. A 180 degree phase shift of the same frequency would be half of a wavelength, or half a mSec of time. At other frequencies, the same time displacement does not create a similar degree of phase shift as other have pointed out. It could be argued that phase is time on a specific frequency by frequency basis only.

The longer the wavelength, (lower frequency) the more time is needed to create that degree of phase shift when compared to an equal phase shift of a shorter wavelength (higher frequency. With that in mind, it makes sense that an offset delay introduced at the crossover’s low output will cause a phase shift that affects the upper end of the bandwidth more so than the lower end of the bandwidth. An equal phase shift across that entire bandwidth would have to be differing delay times for differing frequencies. If I got my money’s worth from Jamie, Sam, Don, and Mr. McCarthy, that would be an all pass filter.

PT

Reply posted by Tom Danley on September 18, 2001
-"An equal phase shift across that entire bandwidth would have to be differing delay times for differing frequencies. If I got my money’s worth from Jamie, Sam, Don, and Mr. McCarthy, that would be an all pass filter.”

This is the behavior of a point source (one who’s diameter is small compared to the wavelength it is producing) with “flat” response. As defined by its electrical equivalent circuit and as measured ala Heyser, a typical woofer, to have flat frequency response, must (mid band) have an acceleration response to the VC force.

This is accomplished by the motor force acting on the drivers moving mass which ends up reflecting an RC filter (C being mass). This 1 pole roll off of the radiator velocity counter acts the improving radiation efficiency (an acoustic/dimension related slope with no phase shift associated with because it is a changing resistance), producing flat response but at about a -90 degree acoustic phase shift (input Voltage with respect to output pressure after all fixed time delays are accounted for).

At the low end, even in a sealed box or infinite baffle, cabinet tuning will cause a large amount of acoustic phase change, going through zero degrees at resonance (Z max where mass and spring are equal but opposite) to a positive value as the system is dominated by compliance stiffness.

Going up, the phase is also zero at the R min point in the impedance, this is where the series L in the VC is equal but opposite the mass reactance (C) and these two terms cancel out leaving the Rdc in series with the acoustic load and losses (a small R).

So you see, you all ready have a “thing” which has a different delay for each frequency, a woofer& most speakers

At low frequencies, unwrapping the acoustic phase back to nominally zero degrees can be done without dsp and when done makes a wonderful sounding subwoofer. Unlike conventional woofers, the zero phase and flat response yields a system which CAN reproduce a complex waveshape.

The normal, non zero acoustic phase is the main thing which has stopped many attempts at active sound cancellation in its simplest form. I have spent a great deal of time working on speakers which had as little acoustic phase change over the widest frequency range possible as well and would also say that makes a significant audible difference.

On the other hand, I do everything with drivers, crossovers and horns and physical placement, partly because I want to actually attack the real problem but also because I am not to hip actually working with dsp.

I know it is possible to correct all the phase stuff this way too and there is at least one hifi dsp correction product which claims to do this, at least at the microphone location.

This is one area where an efficient horn can have an edge, to the extent they are dominated by the acoustic load, a resistance, there acoustic phase is resistive about zero degrees (output pressure and input voltage coincide over a wide range of frequencies).

Cheers,
Tom Danley

Reply posted by Chip on September 19, 2001
Tom,

Two questions:

1) would you consider preparing a “idiots guide to LF phase”? I’m very interested in the phase artifacts caused by different types of boxes and venting / porting methods. I never realy considered this as such a huge factor in the differing performance of different types of boxes.

2) How would you best describe an all-pass filter, and it’s effects?

David, et all, please respond as well. This type of shared information is why we are here.

Thanks,
Chip

Reply posted by Nathan Butler on September 18, 2001
Whenever I think of this, I like to reference some simple mathematics. Bear with me…

A pure tone can be described by:

1) cos(wt) where w = frequency, t = time

With a phase shift (1) becomes:

2) cos(wt + p) where p = phase in degrees

With a time delay, (1) becomes:

3) cos(w(t + td)) where td = time delay

As an example, let’s say w = 100, p = 90, and td = 0.9

(2) becomes cos(100t + 90)

(3) also becomes cos(100t + 90)

Now let’s make the frequency, w = 200

(2) becomes cos(200t + 90)

(3) becomes cos(200t + 180)

Essentially, a time delay yields similar results to a phase shift, except that a time delay increases the phase shift with frequency.

Hope this helps.
Nathan Butler

Reply posted by Patrik Arnekvist on September 18, 2001
I gotta go and try if changing phase angle on my omnidrive to, say 90 deg changes the delay, I guess it will. but how does one come to the conclusion that here I need to shift the phase angle 90 deg?

I can´t really inderstand why one would need to do that.

But changing the delay must be different, cause the time of 90 deg at 5kHz is quite different than 90deg at 100 Hz..right?

I´m getting a headache here, excuse me for thinking out loud.

Patrik

Reply posted by Chip on September 18, 2001
Patrik,

If I understand the electronics correctly, the phase adjustment in the DSP simply “rolls” the phase without adjusting time.

If I’m correct, changing the phase would not change the result of an impulse time measurement.

To the best I can understand it, at this point, the differentiating factor between the requirement for a phase adjustment VS. a time adjustment would be the phase angle relationship AT the crossover frequency. If the angles are not parallel, some time adjustment is required.

Once the phase angles are parallel, if necessary, you can adjust the phase to make the phase angles lay one on top of the other while remaining parallel. If you were to continue to adjust the time, you would loose the coherance of the phase angles and have a less phase coherant area above, and below, the crossover point.

Reply posted by David Gunness on September 18, 2001
Everything you’ve said is true - so I think you’ve got it. Let’s see if I can clarify it.

If the arrival times are the same, the phase response curves will be parallel. If the two signal paths are “in phase” at crossover, the phase response curves will have the same value at the crossover frequency. The ideal situation is for both of these conditions to be true, but occasionally you can’t achieve both by only adjusting delay.

David Gunness

Reply posted by Chip on September 18, 2001
David,

This is exactly what I was getting at. You were able to deliver it in a much more understandable way than I was.

Can you think of any other way to differentiate between the two?

Chip

Reply posted by David Gunness on September 18, 2001
Delay always produces phase shift, but the phase response can be shifted without producing delay.

Dave